Problem 33 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by \(y=x^{2}, x=1,\) and \(y=0 .\) Use the shell method to find the volume of the solid generated when \(R\) is revolved about the following lines. $$x=-2$$

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{7}{6} \pi\) cubic units.

Step 1: Sketch the graph and find the region R

A quick sketch of \(y = x^2,\) \(x = 1,\) and \(y = 0\) will help us visualize the region R and the solid generated by revolving it around the line \(x = -2\). Note that the region is bound on the left by the line \(x = 1\), on the right by the y-axis, and on the bottom by the x-axis.

Step 2: Set up the shell method formula

Now, we'll set up the volume integral using the shell method formula. In this case, the bounds of integration are \(a = 0\) and \(b = 1\). The radius \(p(x)\) is the distance from the line \(x = -2\) to the shell, which is \(x + 2\). The height \(h(x)\) is equal to the function value, which is \(x^2\). Therefore, the volume integral is: $$V = 2 \pi \int_0^1 (x+2)x^2 dx$$

Step 3: Evaluate the integral

Next, we'll evaluate the integral to find the volume of the solid: $$V = 2 \pi \int_0^1 (x+2)x^2 dx = 2\pi \int_0^1 (x^3 + 2x^2)dx$$ Now we can integrate: $$V = 2\pi \left[\frac{1}{4}x^4 + \frac{2}{3}x^3\right]_0^1$$ Now, we'll plug in the bounds: $$V = 2\pi \left(\frac{1}{4}(1)^4 + \frac{2}{3}(1)^3 - \frac{1}{4}(0)^4 - \frac{2}{3}(0)^3\right)$$

Step 4: Simplify and find the volume

Finally, we'll simplify and find the volume of the solid: $$V = 2\pi \left(\frac{1}{4} + \frac{2}{3}\right) = 2\pi \left(\frac{7}{12}\right) = \frac{7}{6} \pi$$ Thus, the volume of the solid generated by revolving the region \(R\) around the line \(x = -2\) is \(\frac{7}{6} \pi\) cubic units.

Key Concepts

Volume of Solids of Revolution

The concept of the volume of solids of revolution involves creating a 3-D shape by rotating a 2-D region around a specific axis. Imagine a flat shape, like a piece of paper, being spun around a pencil. This rotation creates a volume, just as spinning a doughnut makes it appear solid. In our example, we have a region defined by the curve \(y = x^2\), the line \(x = 1\), and the \(x\)-axis. This shape begins looking like a flat slice, but when we rotate it around the line \(x = -2\), it becomes a full three-dimensional object.

By understanding this rotation process, you grasp how an initially flat region grows into a volume. The Shell Method is often used in conjunction with this concept to calculate the new volume formed by such rotations. Keep in mind that not all rotation axes are direct; sometimes, as with \(x = -2\), they are not just the usual \(x\)- or \(y\)-axes, which adds a layer of complexity to the problem.

Integration Techniques

Integration techniques are vital mathematical tools used to find solutions in calculus, helping us determine areas, volumes, and other quantities. The Shell Method in particular is a technique used to find volumes of rotated solids, and it's especially useful for shapes with a cylindrical shell form. This method involves setting up an integral that represents the volume of these shells.

Each shell is a thin slice of the solid, and when considered over the interval from \(a\) to \(b\), the heights of these shells are given by the function defining the 2-D region. For our example, the height \(h(x)\) is \(x^2\), and the radius of each shell is determined by the distance to the axis of rotation \(x=-2\), calculated as \(x+2\). You then calculate the integral: \(2 \pi \int_a^b p(x)h(x)\,dx\), which yields the total volume.

Definite Integrals

Definite integrals take us from dealing with the abstract idea of area under a curve to calculating exact numerical values, which is an essential aspect of calculus. In the context of finding volumes via the Shell Method, a definite integral helps consolidate all the tiny cylindrical shells into a final volume. The definite integral is evaluated over a specified interval along the axis of revolution.

Here, once we set the integral \(2 \pi \int_0^1 (x+2)x^2 \, dx\), definite integration converts this into a calculable number by plugging in bounds 0 and 1. This process smoothly shifts from an abstract sum to an actual number: \(\frac{7}{6} \pi\) cubic units, as found in our problem. This indicates that a definite integral efficiently provides the volume of the solid transformed through rotation, converting infinite little sums into tangible results.